$P(P(A_1-${-1}$))$ where $A_1=${-1,0,1}
I interpret this question as "Find the power set of the power set of $A_1$-{-1}."
Assuming $A_1$-{-1}={0,1}=$B$ (I think this assumption is correct? It could be interpreted as $A_1$- the set of -1, which might just be $A_1$?)
$P(B)$={{0},{1},{0,1}, empty set}
$P(P(B))$={{0},{1},{0,1}, empty set}
Because sets can't have duplicates and ordering does not matter in sets?
Given a set $D$, $P(D)$ has $2^{|D|}$ elements. Hence your computation of $P(P(B))$ is not correct.
$$P(B)=\{\{0\}, \{1\}, \{0,1\}, \emptyset\}$$
For each element of $P(B)$, we need to decide whether it is in a subset or not.
That is
\begin{align} P(P(B)) &= \{ \\&\emptyset, \{\{0\}\}, \{\{1\}\} , \{\{0,1\}\}, \{\emptyset\}, \\ &\{\{0\}, \{1\}\}, \{\{0\}, \{0,1\}\}, \{\{0\}, \emptyset\}\},\\ &\{\{1\}, \{0,1\}\}, \{\{1\}, \emptyset\}\}, \{\{0,1\}, \emptyset\},\\ &P(B)-\{\{0\}\},P(B)-\{\{1\}\}, P(B)-\{\{0,1\}\}, P(B) - \{\emptyset\}, P(B)\\ \} \end{align}
Note that there are exactly $16$ elements. I have expressed the last few terms in terms of $P(B)$ but you should write it out explicitly as an exercise.